If $e=\lim\limits_{x\to \infty} (1+x^{-1})^x$, why doesn't $e=1$? I'm sure that this is a very basic question, but it has been bothering me for a while:
If $e=\lim\limits_{x\to \infty} (1+x^{-1})^x$, shouldn't $e=1$?
If $x$ is tending towards infinity, why wouldn't $x^{-1}$ tend to $0$ and $e=1^\infty=1$?
Thanks
 A: Try thinking about the limit as $x$ goes to infinity of $(1/x)\cdot x$.
Your reasoning tells me that as $x$ gets large, $(1/x)$ gets close to $0$, so $(1/x)\cdot x$ gets close to $0\cdot x$, which is $0$.
Obviously, that's the wrong answer, though, since $(1/x)\cdot x$ is always equal to 1, and so must go to $1$ as $x$ goes to infinity.
Moral:  If part of your expression is getting smaller and the other part is getting bigger, you can't just look at each part separately.  You've got to think about how they interact.
A: Your reasoning basically says
\begin{align}
\lim_{n \to \infty} \lim_{k \to \infty} (1 + \frac{1}{k})^n
=\lim_{n \to \infty} (1 + 0)^n 
=\lim_{n \to \infty} (1)^n 
= 1
\end{align}
which is correct.
But let me note that 
\begin{align}
\lim_{k \to \infty} \lim_{n \to \infty} (1 + \frac{1}{k})^n
=\lim_{k \to \infty} \infty 
=\infty.
\end{align}
If you take either of the individual limits first, you get two quite different answers, and neither of them matches the answer you get when you take them simultaneously. A little numerical experiment with $n = 20$, say, will convince you of this. 
In fact, if you look at the binomial expansion of $(1 + \frac{1}{n})^n$, for $n > 1$, the first two terms are
$$
(1 + \frac{1}{n})^n \approx 1^n + n \cdot 1^{n-1} (\frac{1}{n})^1 + \ldots \ge 2
$$
so the limit clearly must be greater than 2 as well. 
A: You are interpreting $$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x$$ 
as 
$$\lim_{x\to\infty} \left(\lim_{y\to\infty} 1 + \frac{1}{y}\right)^x.$$
Whereas this latter limit is 1, it is not equal to the former limit. 
You cannot take the limit in steps; you have to let $x$ go to infinity simultaneously everywhere it appears in the expression.
To see why $e > 1$, just consider the numerical values of the quantities 2, $(1+\frac{1}{2})^2$, $(1+\frac{1}{3})^3$, etc.
A: Implicitly you are assuming that the limit of $x$ within the bracket and that outside the bracket can be applied separately. That's where you are wrong. It should be applied simultaneously. Basically your argument implies,

$$\displaystyle\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=\left(\displaystyle\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\right)^{\left(\displaystyle\lim_{n\to\infty}n\right)}$$

But it is precisely where you are wrong.
A: Intuitively, you are looking at
$$ (\text{something slightly bigger than } 1)^{\text{something going to } \infty} $$
Now, the base number is $1 + 1/n$ and it becomes closer and closer to $1$ as $n\to\infty$, with speed $n$. At the same time, the exponent tends to $\infty$ with speed $n$.
You know that for any (finite and positive) number $x$ you have $x^\infty = \infty$ and $1^x = 1$, but our situation is subtle, since the base is just tending to $1$ and the exponent is just tending to infinity. 
You kind of have two competing forces pulling a rope and the question is: who is going to win?
Surprisingly there's no winner, but an equilibrium value: the famous $e$. For this reason, I like to think that $e$ stands for equilibrium. 
A: Roughly speaking: By the binomial theorem $$\eqalign{e&:=\lim_{x\to\infty}\left(1+\dfrac1x\right)^x\\&\color{white}:=\lim_{x\to\infty}
    {x \choose 0}1^x \left(\dfrac1x\right)^0 + {x \choose 1}1^{x-1}\left(\dfrac1x\right)^1 + {x \choose 2}1^{x-2}\left(\dfrac1x\right)^2 + \cdots + {x \choose x-1}1^1 \left(\dfrac1x\right)^{x-1} + {x \choose x}1^0 \left(\dfrac1x\right)^x, }$$ as $x\to\infty$, those binomial coefficients takes on a large value, and the $(\tfrac1x)^n$ terms takes on extremely small values, so their product gives us some “small stuff” that when summed up converges to some value. And we know for sure that this value is at least $2$ since the sum of the two first terms in the binomial expansion is equal to $2$.
A: Hint
Consider $$ A=\left(1+\frac{1}{x}\right)^x$$ and take the logarithms of both sides to get $$\log(A)={x}\log\left(1+\frac{1}{x}\right)$$ When $x$ is very large, using Taylor $$\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$$ So $$\log(A)=1-\frac{1}{2 x}+\frac{1}{3 x^2}+\cdots$$ Now, take the limit.
A: Because at the same time as you are making the $p$ in $p^x$ closer to one, you are raising it to a higher and higher poser $x$.
A simpler case of the same sort of effect: What do you make of 
$$\lim_{x\rightarrow \infty} \frac{(4/x) + (1/x^2)}{(4/x)}
$$
One way to look at it is that the numerator approaces zero, so why wouldn't this be zero?
But another, more correct,  way of looking at is to notice that
$$\lim_{x\rightarrow \infty} \frac{(4/x) + (1/x^2)}{(4/x)}
= \lim_{x\rightarrow \infty} \frac{4 + (1/x)}{4} = \lim_{x\rightarrow \infty} (1 + (1/4x) = 1
$$
Here the denominator was shrinking as fast as the numerator.  There, the power was increasing about "as fast as" the thing being exponentiated was approaching 1.
A: Consider the following
$$\begin{array}{lll}
\frac{de^t}{dt}&=&\lim_{x\to\infty}\frac{e^{t+\frac{1}{x}}-e^t}{\frac{1}{x}}\\
&=&e^t\lim_{x\to\infty}\frac{e^\frac{1}{x}-1}{\frac{1}{x}}\\
&=&e^t\lim_{x\to\infty}\frac{((1+\frac{1}{x})^x)^\frac{1}{x}-1}{\frac{1}{x}}\\
&=&e^t\lim_{x\to\infty}\frac{1+\frac{1}{x}-1}{\frac{1}{x}}\\
&=&e^t\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{x}}\\
&=&e^t\\
\end{array}$$
I know that my math is a bit "off", but it does suggest that $e\ne 1$ because $\dfrac{de^t}{dt}=\dfrac{d1^t}{dt}=\dfrac{d1}{dt}=0$, and $\dfrac{de^t}{dt}=e^t= 1^t=1$ implies that $1=0$, which is ridiculous (at least in Real and Complex numbers).
